Let tangent BC touch the circle at point R

As we know tangents drawn from an external point to a circle are equal.

We have

AP = AQ

[tangents from point A]

BP = BR …[1]

[tangents from point B]

CQ = CR …[2]

[tangents from point C]

Now,

AP = AQ

⇒ AB + BP = AC + CQ

⇒ 5 + BR = 6 + CR [From 1 and 2]

⇒CR = BR - 1 …[3]

Now,

BC = 4 cm

BR + CR = 4

BR + BR - 1 = 4 [Using 3]

2BR = 5 cm

BR = 2.5 cm

BP = BR = 2.5 cm [Using 2]

AP = AB + BP = 5 + 2.5 = 7.5 cm